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Define G(x) = Sum_{n>=0} a(n)*x^n/2^[n*(n-1) - A000120(n)], then [x^n] G(x)^(1/2^n) = 1 for n>=0, where A000120(n) = number of 1's in binary expansion of n.
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%I #3 Mar 30 2012 18:37:07

%S 1,2,11,247,87453,30392377,83081803051,447717938403725,

%T 76261525038193025805,6426287262393575837153381,

%U 4292008745048222678362226977889,5685934933249315447199351722237681091

%N Define G(x) = Sum_{n>=0} a(n)*x^n/2^[n*(n-1) - A000120(n)], then [x^n] G(x)^(1/2^n) = 1 for n>=0, where A000120(n) = number of 1's in binary expansion of n.

%F Limit_{n->infinity} Sum_{k=0..n-1} [x^k] G(x)^(1/2^n) = 2, where G(x) satisfies [x^n] G(x)^(1/2^n) = 1 for n>=0.

%e This sequence forms the numerators of coefficients in G(x), which begin:

%e [1,2,11/2,247/16,87453/2048,30392377/262144,83081803051/268435456,...].

%e The denominators are 2^b(n) where b(n) takes on the values:

%e [0,0,1,4,11,18,28,39,55,70,88,107,130,153,179,206,239,270,304,339,...]

%e which is b(n) = n*(n-1) - A000120(n) for n>1 with b(0)=b(1)=0.

%e Illustrate [x^n] G(x)^(1/2^n) = 1 for n=0..5 by:

%e G(x) = (1)+2x +11x^2/2 +247x^3/2^4 +87453x^4/2^11 +30392377x^5/2^18 +...;

%e G(x)^(1/2) = 1 +(x)+9x^2/2^2 +175x^3/2^5 +54685x^4/2^12 +16941497x^5/2^19 +..;

%e G(x)^(1/4) = 1 +x/2 +(x^2)+143x^3/2^6 +41437x^4/2^13 +119466176x^5/2^20 +...;

%e G(x)^(1/8) = 1 +x/2^2 +15x^2/2^5 +(x^3)+35541x^4/2^14 +9826265x^5/2^21 +...;

%e G(x)^(1/16) = 1 +x/2^3 +29x^2/2^7 +483x^3/2^10 +(x^4) +8853753x^5/2^22 +...;

%e G(x)^(1/32) = 1 +x/2^4 +57x^2/2^9 +1875x^3/2^13 +251395x^4/2^19 +(x^5)+...;

%e so that the coefficient of x^n in G(x)^(1/2^n) equals 1 for n>=0.

%e To illustrate that the n-th partial sums of G(x)^(1/2^n) approaches 2: at n=5, Sum_{k=0..4} [x^k] G(x)^(1/32) = 1+1/2^4+57/2^9+1875/2^13+251395/2^19 = 1.88...

%o (PARI) {a(n)=local(A=[1]);if(n==1,2, for(i=0,n,A=Vec(Ser(concat(Vec(Ser(A)^(1/2^#A)),1))^(2^#A))); A[n+1]*2^(n*(n-1))/2^subst(Pol(binary(n)),x,1))}

%Y Cf. A000120.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Oct 26 2007, Oct 29 2007